two cards are dealt in succession from a standard deck of cards. what is the probability that the second card is red give that the first card was a heart. so i do p(R/H), then i know you go R*H/H. but how do i get the probability please explain! thanks.
if 2 dice are rolled, what is the probability of rolling at least one three?
i nkow that i have to use binomial probablity but how do i get the answer?
how do i simplify (x-2)!/x! i know that (x-2)! is (x-2)(x-1) but what is x! factorial? thanks for your time!
first card was red
so now I have 51 cards
25 are red and 26 are black
so
25/51
( are you sure you typed this question the exact way it was in the book?)
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probability of the first 3 = 1/6 and the second not 3 = 1/6*5/6 = 5/36
since probability that the first was not 3 =5/6
probability that the second was 3 and the first not 3 = 1/6*5/6 = 5/36
probability that both are 3 = 1/6*1/6
sum = 1/6 + 5/36 + 5/36 = 12/36 = 11/36
OK, using BINOMIAL
p = 1/6
1-p = 5/6
P(n,k) = C(n,k) p^k (1-p)^(n-k)
we want P(2,1) + P(2,1)
C(2,1) = 2
C(2,2) = 1 from formula or Pascal triangle
P(2,1) = 2*(1/6)^1 *(5/6)^1
P(2,2) = 1*(1/6)^2 *(5/6)^0
SUM = 2*5/36 + 1/36 =11/36
I did it both ways so you can study what is really going on.
probability of the first 3 = 1/6 and the second not 3 = 1/6*5/6 = 5/36
since probability that the second was not 3 =5/6
probability that the second was 3 and the first not 3 = 1/6*5/6 = 5/36
probability that both are 3 = 1/6*1/6
sum = 1/6 + 5/36 + 5/36 = 11/36
OK, using BINOMIAL
p = 1/6
1-p = 5/6
P(n,k) = C(n,k) p^k (1-p)^(n-k)
we want P(2,1) + P(2,1)
C(2,1) = 2
C(2,2) = 1 from formula or Pascal triangle
P(2,1) = 2*(1/6)^1 *(5/6)^1
P(2,2) = 1*(1/6)^2 *(5/6)^0
SUM = 2*5/36 + 1/36 =11/36
(x-2)!/x! i know that (x-2)! is (x-2)(x-1)
(x-2)(x-3)((x-4) .....
------------------------
x (x-1)(x-2)(x-3) (x-4) .....
= 1/ [ x(x-1)]
i still don't under the simplifying one cause reiny told me that (x+2)! was (x+2) (x+1) but howcome you are telling me that you go up in factorial. i though factorial goes down
To find the probability that the second card is red given that the first card was a heart, you can use the definition of conditional probability. The probability notation you mentioned, P(R/H), represents the probability of event R (second card is red) given event H (first card is a heart).
To find this probability, you need to divide the number of favorable outcomes by the total number of possible outcomes. Here's how you can calculate it:
1. Determine the number of favorable outcomes.
- Assuming the first card is a heart, there are 26 cards left that are red (i.e., 26 red cards out of the remaining 51 cards).
2. Determine the total number of possible outcomes.
- Since the first card is already chosen and is a heart, there are now 51 cards remaining to choose from.
3. Calculate the probability.
- Probability (P(R/H)) = Number of favorable outcomes / Total number of possible outcomes
- P(R/H) = 26/51
So, the probability that the second card is red given that the first card was a heart is 26/51.
For the second question about rolling two dice and getting at least one three, you can use the principle of complementary probability. Instead of calculating the probability of rolling at least one three directly, you can calculate the probability of not rolling any threes and then subtract that from 1.
To calculate this:
1. Determine the probability of not rolling any threes on one die.
- There are 6 possible outcomes on a fair die, and only 1 of them is a three. So, the probability of not rolling a three on one die is 5/6.
2. Since you have two dice, the probability of not rolling any threes on both dice is (5/6)^2 = 25/36.
3. Calculate the probability of rolling at least one three.
- This can be found by subtracting the probability of not rolling any threes from 1.
- Probability (At least one three) = 1 - Probability (No threes)
- Probability (At least one three) = 1 - 25/36
- Probability (At least one three) = 11/36
Therefore, the probability of rolling at least one three when rolling two dice is 11/36.
Regarding simplifying the expression (x-2)!/x!, you can simplify it using the properties of factorial.
1. Recall the definition of factorial, n!, which represents the product of all positive integers from 1 to n.
- For example, 4! = 4x3x2x1 = 24
2. Simplify the expression (x-2)!
- (x-2)! = (x-2)(x-3)(x-4)...(2)(1)
3. Simplify the expression x!
- x! = x(x-1)(x-2)(x-3)...(2)(1)
4. Divide (x-2)! by x!
- (x-2)!/x! = [(x-2)(x-3)(x-4)...(2)(1)]/[x(x-1)(x-2)...(2)(1)]
Notice that the (x-2) terms cancel out in the numerator and denominator, leaving:
- (x-2)!/x! = 1/(x(x-1))
So, the simplified form of (x-2)!/x! is 1/(x(x-1)).